Romanian Mathematical Olympiad

Question

For every positive, odd integer n, prove that [ 12 + n + 12 ] = [ 12 + n + 12020 ], where [a] denotes the integer part of the real number a.

Accepted Answer

We will prove by contradiction that there is no integer between

\frac{1}{2} + n + \frac{1}{2020}

and

\frac{1}{2} + n + \frac{1}{2}

.

Suppose there exists

k \geq 1

such that

\frac{1}{2} + n + \frac{1}{2} \geq k > \frac{1}{2} + n + \frac{1}{2020}

. We obtain

n \geq k^{2} - k - \frac{1}{4} = x_{k}

and

n < k^{2} - k + \frac{126}{505} = y_{k}

.

Since the only integer between

x_{k}

and

y_{k}

is

k^{2} - k

, we must have

n = k^{2} - k = k (k - 1)

which contradicts

n

odd.